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A poll conducted in 2013 found that \(52 \%\) of U.S. adult Twitter users get at least some news on Twitter. \(^{59}\) The standard error for this estimate was \(2.4 \%\), and a normal distribution may be used to model the sample proportion. Construct a \(99 \%\) confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

Short Answer

Expert verified
The 99% confidence interval is 45.82% to 58.18%.

Step by step solution

01

Identify the given values

From the question, we know the sample proportion \( \hat{p} = 0.52 \), the standard error \( SE = 0.024 \) (since 2.4% is converted to a decimal), and we are asked to construct a 99% confidence interval.
02

Determine the critical value

The critical value for a 99% confidence interval using the normal distribution (Z-distribution) is approximately 2.576. This value corresponds to the Z-score that captures the central 99% of the distribution.
03

Calculate the margin of error (ME)

Compute the margin of error using the formula: \( ME = Z_{score} \times SE \). Substitute the values: \( ME = 2.576 \times 0.024 \approx 0.0618 \).
04

Construct the confidence interval

The 99% confidence interval is constructed using the formula: \( \hat{p} \pm ME \). So, the interval is: \( 0.52 \pm 0.0618 \). This calculates to: \( 0.4582 \) to \( 0.5818 \).
05

Interpret the confidence interval

We are 99% confident that the true proportion of U.S. adult Twitter users who get at least some news on Twitter falls between 45.82% and 58.18%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion is a key concept when dealing with confidence intervals and statistical inference. Imagine you conducted a survey among a group of people to know how many of them get news from Twitter. The sample proportion is simply the result of your survey. It's the fraction or percentage of the sample responding in a certain way. In our context, the sample proportion \( \hat{p} \) is 0.52 or 52% of U.S. adult Twitter users. This means that out of all the people surveyed, 52% reported getting their news from Twitter.

To calculate it, take the number of people in your sample that expressed the characteristic (getting news from Twitter) and divide it by the total size of the sample. Sample proportions help us predict how the entire population might behave based on a smaller, manageable group of observations.
Standard Error
The standard error (SE) is a statistical measure that tells us how much the sample proportion is expected to vary across different samples in a population. It's crucial for understanding the precision of your sample estimate. In simpler terms, it's a way to measure the spread or variability of the sample data.

In our exercise, the standard error is given as 0.024 (or 2.4%). This small value of SE implies that if we were to take multiple samples of similar size, the proportions derived from them would typically deviate by 2.4% from the true population proportion. The formula to calculate SE for a proportion is: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]where \( \hat{p} \) is the sample proportion and \( n \) is the sample size. A lower standard error indicates a more reliable sample estimate.
Normal Distribution
Normal distribution is a continuous probability distribution that is symmetric around the mean. It's often visualized as the bell curve. In statistics, it’s standard to model the distribution of sample means or proportions using this distribution because of the Central Limit Theorem.

In the context of confidence intervals, the normal distribution allows us to use Z-scores to create these intervals. For our exercise, it helps us understand how the sample proportions are expected to behave in repeated sampling. Because we assume a normal distribution, we can use the critical value associated with the Z-distribution, in our case 2.576, to calculate the margin of error for a 99% confidence interval. This means we're calculating a range that should capture the true population proportion 99 times out of 100.
Margin of Error
The margin of error is a measure of the uncertainty associated with the sample proportion. It provides a range that the true population parameter is expected to fall within. In our exercise, we've calculated the margin of error, using the formula: \[ ME = Z_{score} \times SE \]where the critical Z-score is 2.576 and the standard error is 0.024. This resulted in a margin of error of approximately 0.0618 or 6.18%.

The margin of error is added and subtracted from the sample proportion to build a confidence interval. This range, therefore, gives us a sensible estimate for the true proportion of U.S. adult Twitter users who obtain news from the platform. Large margins indicate less certainty and precision, while smaller margins suggest greater confidence in the sample's representation of the population.

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Most popular questions from this chapter

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. \(^{56}\) (a) What percent of passenger vehicles travel slower than 80 miles/hour? (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (c) How fast do the fastest \(5 \%\) of passenger vehicles travel? (d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

Find the standard deviation of the distribution in the following situations. (a) MENSA is an organization whose members have IQs in the top \(2 \%\) of the population. IQs are normally distributed with mean 100 , and the minimum IQ score required for admission to MENSA is 132 . (b) Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\). Women with cholesterol levels above \(220 \mathrm{mg} / \mathrm{dl}\) are considered to have high cholesterol and about \(18.5 \%\) of women fall into this category.

A poll conducted in 2013 found that \(52 \%\) of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was \(2.4 \%\). Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. (a) The data provide statistically significant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of \(\alpha=0.01\). (b) Since the standard error is \(2.4 \%\), we can conclude that \(97.6 \%\) of all U.S. adult Twitter users were included in the study. (c) If we want to reduce the standard error of the estimate, we should collect less data. (d) If we construct a \(90 \%\) confidence interval for the percentage of U.S. adults Twitter users who get some news through Twitter, this confidence interval will be wider than a corresponding \(99 \%\) confidence interval.

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